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Re: [EMHL] Fermat Points

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  • Steve Sigur
    Each of these triangles has a point U = midpoint of GH. The Fermat line goes through this point, so either the U points of the three triangles involved are
    Message 1 of 10 , Dec 8, 2005
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      Each of these triangles has a point U = midpoint of GH. The Fermat
      line goes through this point, so either the U points of the three
      triangles involved are colinear, or they are the same.

      Steve


      On Dec 7, 2005, at 9:22 AM, Wilson Stothers wrote:

      > Dear All,
      >
      > We have the following fact :
      >
      > If F is a Fermat Point of triangle ABC, then F is also a
      > Fermat Point of its Cevian, and of its Anticevian Triangle.
      >
      > A couple of questions :
      >
      > Is this well-known? I'd appreciate a reference.
      >
      > Apart from G, are there other centres with this property?
      >
      > I also observe from Cabri that, if F is a Fermat point,
      > and if ABC and the Cevian and Anticevian Triangles have
      > Fermat Points {F,F2},{F,F3},{F,F4}, respectively, then
      > (a) the points F,F2,F3,F4 are concyclic,
      > (b) the circle also contains the isogonal conjugates of
      > F2, F3, F4 with respect to their triangles,
      > (c) the point F and its isogonal conjugates in the three
      > triangles are collinear,
      > (d) the Euler Lines of the three triangles are parallel,
      > this is an easy consequence of (c).
      >
      > Of course the isogonal conjugates in (b) and (c) are the
      > Isodynamic Points of their triangles. Thus, if F = X13,
      > then the circle is that through X13, X14 and X16, and
      > the line is X13X15. Note that F need not be X13 for the
      > other two triangles, but is a Fermat Point.
      >
      > I suspect that (c) characterizes the Fermat Points, and
      > perhaps that (d) also does.
      >
      > I'd be glad of any proofs or comments on these. I think
      > that a circle with seven points is unusual. The centre
      > might be interesting, it is on X115X125 as this is the
      > perpendicular bisector of X13X14.
      >
      > Seasonal Greetings
      >
      > Wilson
      >
      >
      >
      >
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    • Wilson Stothers
      Dear Bernard, I know nothing about angular coordinates, so hesitate to say much Your proof below suggests that, if F = X13, then it is also X13 of its cevian
      Message 2 of 10 , Dec 8, 2005
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        Dear Bernard,

        I know nothing about angular coordinates, so hesitate to say much
        Your proof below suggests that, if F = X13, then it is also X13 of
        its cevian and anticevian triangle. This isn't always so, it can
        be X14 for one or other.


        > > We have the following fact :
        > >
        > > If F is a Fermat Point of triangle ABC, then F is also a
        > > Fermat Point of its Cevian, and of its Anticevian Triangle.
        >
        > this is direct consequence of the fact that the angular
        coordinates
        > of a point wrt ABC, its cevian and anticevian triangles are the
        same.

        Is this true?
        I'm guessing at the definition, but I think there may be
        supplements involved. Probably I just guessed wrongly.

        > for X13 : 120°, 120°, 120°
        >
        > [snip]
        >
        >

        Yours

        Wilson
      • Steve Sigur
        Ah, that is what I wanted to know. But it seems to me that their angular coordinates would not change, so both points should be the same. It puzzles me that
        Message 3 of 10 , Dec 8, 2005
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          Ah, that is what I wanted to know. But it seems to me that their
          angular coordinates would not change, so both points should be the
          same. It puzzles me that the second points are different. Intriguing
          if they are.

          Oh I know, it must be true that if we get the pedal and antipedal of
          Fs, then the three Fs's are the same. This makes 12 points from these
          six triangles, which I bet have an interesting relationship, Can't
          wait until I have some time with Sketchpad.

          Hmm, algebra at work.

          Steve


          On Dec 8, 2005, at 9:55 AM, Wilson Stothers wrote:

          > the three triangles will have different second
          > Fermat Points, so the lines FnFs are not parallel.



          [Non-text portions of this message have been removed]
        • Steve Sigur
          Bernard, The two Fermat points each have a cevian and anticevian triangle. The 4 Euler lines of these triangles seem to be parallel to that of ABC (observed on
          Message 4 of 10 , Dec 9, 2005
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            Bernard,

            The two Fermat points each have a cevian and anticevian triangle. The
            4 Euler lines of these triangles seem to be parallel to that of ABC
            (observed on Sketchpad, not proved). Is there a curve, all of whose
            points have Euler line parallel to that of ABC?

            Steve
          • Bernard Gibert
            Dear Steve, ... It is true. ... this was answered by Wilson yesterday under the thread [EMHL] The Cubic K278! Best regards Bernard [Non-text portions of this
            Message 5 of 10 , Dec 9, 2005
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              Dear Steve,

              > The two Fermat points each have a cevian and anticevian triangle. The
              > 4 Euler lines of these triangles seem to be parallel to that of ABC
              > (observed on Sketchpad, not proved).

              It is true.


              > Is there a curve, all of whose
              > points have Euler line [of their cevian and anticevian triangles, I
              > suppose] parallel to that of ABC?

              this was answered by Wilson yesterday under the thread [EMHL] The
              Cubic K278!


              Best regards

              Bernard



              [Non-text portions of this message have been removed]
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